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(And if so, how can I describe the "multiplication" on the sequence?)

We consider a bijection, denoted by $f$, from the algebraic closure of $\mathbb{F}_2$, named $\bar{\mathbb{F}_2}$, to the set $S\subset\{(b_j)_{j=-\infty}^{\infty} : b_j \in \{0,1\}\}$ which contains all the periodicly repeating 0-1 sequences, or to say, $S:=\{(b_j)_{j=-\infty}^{\infty} : b_j \in \{0,1\}, \exists T \in \mathbb{N}^+, \forall j \in \mathbb{Z}, b_j = b_{j+T}\}$. Denote the k-th entry of $f(x)\in S$ by $f_k(x)$. We hope the bijection to fulfill the properties below:

  1. The addition on $\bar{\mathbb{F}_2}$ is mapped onto the bitwise xor on $S$, or to say, for all $x,y \in \bar{\mathbb{F}_2}$ and $k \in \mathbb{Z}$, $f_k(x+y) = f_k(x) \oplus f_k(y)$ where $\oplus$ represents the exclusive or.
  2. For any nonzero number $x \in \bar{\mathbb{F}_2}$ with minimal polynomial $p(x)$, the period of $f(x)$ is equal to the degree of $p(x)$. Moreover, for any two numbers $x$ and $y$ with the same minimal polynomial, the sequences $f(x)$ and $f(y)$ is identical up to a translation, i.e. exists an integer $t$, for each integer $k$, $f_k(x)=f_{k+t}(y)$.
  3. For any number $x \in \bar{\mathbb{F}_2}$, $f(x^2)$ is $f(x)$ translated one bit right. In other word, $f_{k+1}(x^2)=f_k(x)$.

Now it is known that, for those numbers with minimal polynomial $x^2+x+1$, the sequence is like $(...,1,0,1,0,1,0,1,0,...)$, while for those with minimal polynomial $x^3+x+1$, the sequence is like $(...,1,1,0,1,1,0,1,1,0,...)$.

Now I want to know that, is there a bijection fulfilling the properties above? If so, is it unique up to any automorphism mapping a number onto any number with the same minimal polynomial? And for any instance of the bijection, can we describe the rules of the multiplication on the field briefly, i.e. give an expression for $f(x)$, $f(y)$ and $f(xy)$ for any $x$ and $y$?

Thanks very much!

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  • $\begingroup$ Is your bijection uniquely determined or you mean any bijection with these properties? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 10, 2022 at 10:24
  • $\begingroup$ Unless I missed something, the existence of $f$ is equivalent to the existence of an index-$2$ subgroup $H$ of $(\overline{\mathbb F}_2,+)$ such that every nonzero element of $\overline{\mathbb F}_2$ has a conjugate outside $H$. On the one hand, given $f$, we can define $H=\{x:f_0(x)=0\}$. On the other hand, given $H$, we can define $f$ by $f_k(x)=0\iff x^{2^{-k}}\in H$. It’s a bit tedious, but routine, to check that it satisfies all the required properties. $\endgroup$
    – Emil Jeřábek
    Commented Jun 10, 2022 at 11:02
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    $\begingroup$ Also, a simple compactness argument shows that such an $H$ exists for $\overline{\mathbb F}_2$ iff for all $n$, there exists such an $H$ for $\mathbb F_{2^n}$. $\endgroup$
    – Emil Jeřábek
    Commented Jun 10, 2022 at 11:09
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    $\begingroup$ @EmilJeřábek To see that such an $H$ exists, use the normal basis theorem to find a $y$ such that $y$, $\sigma(y)$, ... $\sigma^{n-1}(y)$ is a basis of $\mathbb{F}_{2^n}$ (where $\sigma$ is the Frobenius map). Then take the kernel of $x \mapsto \text{Tr}(xy)$. If there were some $x$ all of whose conjugates obeyed $\text{Tr}(\sigma^k(x) y) =0$, then we would have $\text{Tr}(x \sigma^{-k}(y)) =0$. But trace is a perfect pairing and the $\sigma^{-k}(y)$ span $\mathbb{F}_{2^n}$, so this implies that $x$ is $0$. $\endgroup$
    – David E Speyer
    Commented Jun 10, 2022 at 11:26
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    $\begingroup$ @DavidESpeyer Rather than use the trace pairing, it's probably slightly simpler to use the dual basis to the normal basis. $\endgroup$
    – Will Sawin
    Commented Jun 10, 2022 at 11:54

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